A bar magnet of mass 0.2 kg is dropped from rest along the axis of a fixed closed coil. After falling through 1 m, its speed is \(4\ \mathrm{m\,s^{-1}}\). Ignore air resistance and take \(g=10\ \mathrm{m\,s^{-2}}\). The heat produced in the coil during this fall is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.4\ \mathrm{J}\)
As the magnet falls, the flux through the coil changes and a current is induced in it. By Lenz’s law, this current opposes the motion of the magnet. So the magnet gains less kinetic energy than the potential energy it loses. The difference appears as heat in the coil.
The potential energy lost is
\( \begin{aligned} mgh &= 0.2\times10\times1 \\ &= 2\ \mathrm{J} \end{aligned} \)
The kinetic energy gained is
\( \begin{aligned} \frac{1}{2}mv^{2} &= \frac{1}{2}\times0.2\times4^{2} \\ &= 1.6\ \mathrm{J} \end{aligned} \)
So the heat produced in the coil is
\( \begin{aligned} H &= mgh-\frac{1}{2}mv^{2} \\ &= 2-1.6 \\ &= 0.4\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Lenz’s Law and Conservation of Energy – “… In this situation, the bar magnet experiences a repulsive force due to the induced current. Therefore, a person has to do work in moving the magnet. … This energy is dissipated by Joule heating produced by the induced current.”